Question

Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition, the number of defects per unit is distributed according to a Poisson distribution. What is the probability that there will be a total of 7 defects on four units

Answer 1

Answer:

The probability that there will be a total of 7 defects on four units  is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let X = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, $(2\times4)=8$.

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.